Let $g$ be a vector-valued function defined by $g(t)=(5t+4,2^{t+1})$. Find $g'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $(5,2^{t+1}\ln(2))$ (Choice B) B $(5t,2^{t+1})$ (Choice C) C $\left(\dfrac52t^2+4t,\dfrac{2^{t+1}}{\ln(2)}\right)$ (Choice D) D $5+2^{t+1}$
Answer: $g$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $g(t)=(5t+4,2^{t+1})$. Let's differentiate the first expression: $\dfrac{d}{dt}(5t+4)=5$ Let's differentiate the second expression: $\dfrac{d}{dt}(2^{t+1})=2^{t+1}\ln(2)$ Now let's put everything together: $\begin{aligned} g'(t)&=\left(\dfrac{d}{dt}(5t+4),\dfrac{d}{dt}(2^{t+1})\right) \\\\ &=(5,2^{t+1}\ln(2)) \end{aligned}$ In conclusion, $g'(t)=(5,2^{t+1}\ln(2))$.